Lower bound for the number of lattice points on high dimensional spheresMany representations as a sum of three squaresAn exact counting solution for the number of points within a circle of radius $r$ centered on a lattice point in a $A_2$ hexagonal latticeLattice points inside a (n-dimensional) tetrahedronBound on the number of lattice points in d-dimensional ballBounding the number of lattice points inside an $n$-dimensional ellipsoidCounting lattice points inside a three-dimensional ellipsoidCan we count the number of integer lattice points in this case?Is there a relation between the number of lattice points lie within these circlesHeuristics behind the Circle problem?Number of lattice points on spheres with center not at the originDivisor bound for $r_2$ off the origin
Lower bound for the number of lattice points on high dimensional spheres
Many representations as a sum of three squaresAn exact counting solution for the number of points within a circle of radius $r$ centered on a lattice point in a $A_2$ hexagonal latticeLattice points inside a (n-dimensional) tetrahedronBound on the number of lattice points in d-dimensional ballBounding the number of lattice points inside an $n$-dimensional ellipsoidCounting lattice points inside a three-dimensional ellipsoidCan we count the number of integer lattice points in this case?Is there a relation between the number of lattice points lie within these circlesHeuristics behind the Circle problem?Number of lattice points on spheres with center not at the originDivisor bound for $r_2$ off the origin
$begingroup$
Let $rS^d-1$ denote the sphere of radius $r$ in dimension $d$ (centered at the origin). I'm interested in the number of lattice points on the sphere (not inside).
More precisely, let $$
N(r,d):=textnumber of lattice points on the sphere of raduis r=#xin rS^d-1: xin mathbbZ^d.
$$
I'm especially interested in the lower bound of $N(r,d)$ for any $dge 3$ and large $r$ (with $r^2inmathbbZ$, of course).
For example, I found in the book by F. Fricker Einführung in die Gitterpunktlehre. (German) [Introduction to lattice point theory] that the following result seems to be true (my German is poor):
$N(r,d)gtrsim r^d-2$ for $dge 4$.
So what about $d=3$ case? What is the current best lower bound? The book is in 1982 so I guess there might be a better exponent than $d-2$ now.
One can also ask a weaker question: is there a sequence of $r$ tending to $infty$ such that the above inequality holds with a better lower bound?
nt.number-theory analytic-number-theory harmonic-analysis lattices
edited 6 hours ago
user64494
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asked 8 hours ago
Tony BTony B
2361 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $rS^d-1$ denote the sphere of radius $r$ in dimension $d$ (centered at the origin). I'm interested in the number of lattice points on the sphere (not inside).
More precisely, let $$
N(r,d):=textnumber of lattice points on the sphere of raduis r=#xin rS^d-1: xin mathbbZ^d.
$$
I'm especially interested in the lower bound of $N(r,d)$ for any $dge 3$ and large $r$ (with $r^2inmathbbZ$, of course).
For example, I found in the book by F. Fricker Einführung in die Gitterpunktlehre. (German) [Introduction to lattice point theory] that the following result seems to be true (my German is poor):
$N(r,d)gtrsim r^d-2$ for $dge 4$.
So what about $d=3$ case? What is the current best lower bound? The book is in 1982 so I guess there might be a better exponent than $d-2$ now.
One can also ask a weaker question: is there a sequence of $r$ tending to $infty$ such that the above inequality holds with a better lower bound?
nt.number-theory analytic-number-theory harmonic-analysis lattices
edited 6 hours ago
user64494
2,1029 silver badges18 bronze badges
asked 8 hours ago
Tony BTony B
2361 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $rS^d-1$ denote the sphere of radius $r$ in dimension $d$ (centered at the origin). I'm interested in the number of lattice points on the sphere (not inside).
More precisely, let $$
N(r,d):=textnumber of lattice points on the sphere of raduis r=#xin rS^d-1: xin mathbbZ^d.
$$
I'm especially interested in the lower bound of $N(r,d)$ for any $dge 3$ and large $r$ (with $r^2inmathbbZ$, of course).
For example, I found in the book by F. Fricker Einführung in die Gitterpunktlehre. (German) [Introduction to lattice point theory] that the following result seems to be true (my German is poor):
$N(r,d)gtrsim r^d-2$ for $dge 4$.
So what about $d=3$ case? What is the current best lower bound? The book is in 1982 so I guess there might be a better exponent than $d-2$ now.
One can also ask a weaker question: is there a sequence of $r$ tending to $infty$ such that the above inequality holds with a better lower bound?
nt.number-theory analytic-number-theory harmonic-analysis lattices
edited 6 hours ago
user64494
2,1029 silver badges18 bronze badges
asked 8 hours ago
Tony BTony B
2361 silver badge12 bronze badges
$endgroup$
Let $rS^d-1$ denote the sphere of radius $r$ in dimension $d$ (centered at the origin). I'm interested in the number of lattice points on the sphere (not inside).
More precisely, let $$
N(r,d):=textnumber of lattice points on the sphere of raduis r=#xin rS^d-1: xin mathbbZ^d.
$$
I'm especially interested in the lower bound of $N(r,d)$ for any $dge 3$ and large $r$ (with $r^2inmathbbZ$, of course).
For example, I found in the book by F. Fricker Einführung in die Gitterpunktlehre. (German) [Introduction to lattice point theory] that the following result seems to be true (my German is poor):
$N(r,d)gtrsim r^d-2$ for $dge 4$.
So what about $d=3$ case? What is the current best lower bound? The book is in 1982 so I guess there might be a better exponent than $d-2$ now.
One can also ask a weaker question: is there a sequence of $r$ tending to $infty$ such that the above inequality holds with a better lower bound?
nt.number-theory analytic-number-theory harmonic-analysis lattices
nt.number-theory analytic-number-theory harmonic-analysis lattices
edited 6 hours ago
user64494
2,1029 silver badges18 bronze badges
asked 8 hours ago
Tony BTony B
2361 silver badge12 bronze badges
edited 6 hours ago
user64494
2,1029 silver badges18 bronze badges
asked 8 hours ago
Tony BTony B
2361 silver badge12 bronze badges
edited 6 hours ago
user64494
2,1029 silver badges18 bronze badges
edited 6 hours ago
user64494
2,1029 silver badges18 bronze badges
edited 6 hours ago
user64494
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2,1029 silver badges18 bronze badges
asked 8 hours ago
Tony BTony B
2361 silver badge12 bronze badges
asked 8 hours ago
Tony BTony B
2361 silver badge12 bronze badges
asked 8 hours ago
Tony BTony B
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2361 silver badge12 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is easy to see that no integer of the form $8k+7$ can be written as the sum of three squares. So there can be no universal lower bound for $N(r,3)$ that is better than $0$. (Indeed the integers not expressible as the sum of three squares have an exact characterization.)
For $dge5$ at least, the circle method as applied to Waring's problem gives essentially an asymptotic formula for $N(r,d)$; more precisely, it gives an asymptotic order of magnitude term times a "singular series" leading constant depending on arithmetic properties of $r^2$, but that leading constant is bounded between two universal positive constants.
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
$endgroup$
$begingroup$
Thanks. Now I see that the exponent $d-2$ can't be improved for $dge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^1-epsilon$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880
$endgroup$
– Tony B
3 hours ago
$begingroup$
@TonyB: For $din3,4$ the quantity is $r^d-2$ up to a factor of $loglog r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details.
$endgroup$
– GH from MO
2 hours ago
add a comment |
$begingroup$
My answer to this MO question contains the answer to your question, especially if you take into account that $Lleft(1,left(fracDcdotright)right)$ can be estimated unconditionally (i.e. without GRH):
$$|D|^-varepsilonll_varepsilon Lleft(1,left(tfracDcdotright)right)ll log|D|.$$
The lower bound is ineffective (i.e. we don't know the implied constant), and it is due to Siegel (1934). The upper bound is effective and older. Both bounds are explained in Montgomery-Vaughan: Multiplicative number theory I.
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is easy to see that no integer of the form $8k+7$ can be written as the sum of three squares. So there can be no universal lower bound for $N(r,3)$ that is better than $0$. (Indeed the integers not expressible as the sum of three squares have an exact characterization.)
For $dge5$ at least, the circle method as applied to Waring's problem gives essentially an asymptotic formula for $N(r,d)$; more precisely, it gives an asymptotic order of magnitude term times a "singular series" leading constant depending on arithmetic properties of $r^2$, but that leading constant is bounded between two universal positive constants.
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
$endgroup$
$begingroup$
Thanks. Now I see that the exponent $d-2$ can't be improved for $dge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^1-epsilon$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880
$endgroup$
– Tony B
3 hours ago
$begingroup$
@TonyB: For $din3,4$ the quantity is $r^d-2$ up to a factor of $loglog r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details.
$endgroup$
– GH from MO
2 hours ago
add a comment |
$begingroup$
It is easy to see that no integer of the form $8k+7$ can be written as the sum of three squares. So there can be no universal lower bound for $N(r,3)$ that is better than $0$. (Indeed the integers not expressible as the sum of three squares have an exact characterization.)
For $dge5$ at least, the circle method as applied to Waring's problem gives essentially an asymptotic formula for $N(r,d)$; more precisely, it gives an asymptotic order of magnitude term times a "singular series" leading constant depending on arithmetic properties of $r^2$, but that leading constant is bounded between two universal positive constants.
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
$endgroup$
$begingroup$
Thanks. Now I see that the exponent $d-2$ can't be improved for $dge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^1-epsilon$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880
$endgroup$
– Tony B
3 hours ago
$begingroup$
@TonyB: For $din3,4$ the quantity is $r^d-2$ up to a factor of $loglog r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details.
$endgroup$
– GH from MO
2 hours ago
add a comment |
$begingroup$
It is easy to see that no integer of the form $8k+7$ can be written as the sum of three squares. So there can be no universal lower bound for $N(r,3)$ that is better than $0$. (Indeed the integers not expressible as the sum of three squares have an exact characterization.)
For $dge5$ at least, the circle method as applied to Waring's problem gives essentially an asymptotic formula for $N(r,d)$; more precisely, it gives an asymptotic order of magnitude term times a "singular series" leading constant depending on arithmetic properties of $r^2$, but that leading constant is bounded between two universal positive constants.
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
$endgroup$
It is easy to see that no integer of the form $8k+7$ can be written as the sum of three squares. So there can be no universal lower bound for $N(r,3)$ that is better than $0$. (Indeed the integers not expressible as the sum of three squares have an exact characterization.)
For $dge5$ at least, the circle method as applied to Waring's problem gives essentially an asymptotic formula for $N(r,d)$; more precisely, it gives an asymptotic order of magnitude term times a "singular series" leading constant depending on arithmetic properties of $r^2$, but that leading constant is bounded between two universal positive constants.
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
answered 7 hours ago
Greg MartinGreg Martin
9,6711 gold badge39 silver badges63 bronze badges
9,6711 gold badge39 silver badges63 bronze badges
$begingroup$
Thanks. Now I see that the exponent $d-2$ can't be improved for $dge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^1-epsilon$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880
$endgroup$
– Tony B
3 hours ago
$begingroup$
@TonyB: For $din3,4$ the quantity is $r^d-2$ up to a factor of $loglog r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details.
$endgroup$
– GH from MO
2 hours ago
add a comment |
$begingroup$
Thanks. Now I see that the exponent $d-2$ can't be improved for $dge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^1-epsilon$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880
$endgroup$
– Tony B
3 hours ago
$begingroup$
@TonyB: For $din3,4$ the quantity is $r^d-2$ up to a factor of $loglog r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details.
$endgroup$
– GH from MO
2 hours ago
$begingroup$
Thanks. Now I see that the exponent $d-2$ can't be improved for $dge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^1-epsilon$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880
$endgroup$
– Tony B
3 hours ago
$begingroup$
Thanks. Now I see that the exponent $d-2$ can't be improved for $dge 5$. I guess $d-2$ is also the best for $d=4$. For $d=3$, there is lower bound $r^1-epsilon$ for certain modulo class of $r^2$. See the first page of arxiv.org/abs/1606.05880
$endgroup$
– Tony B
3 hours ago
$begingroup$
@TonyB: For $din3,4$ the quantity is $r^d-2$ up to a factor of $loglog r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details.
$endgroup$
– GH from MO
2 hours ago
$begingroup$
@TonyB: For $din3,4$ the quantity is $r^d-2$ up to a factor of $loglog r$ both ways. Well, this requires congruence conditions on $r^2$ to hold (e.g. $r^2$ must not be divisible by a high power of $2$), and for $d=3$ the best bounds are conditional under GRH. See my answer below for more details.
$endgroup$
– GH from MO
2 hours ago
add a comment |
$begingroup$
My answer to this MO question contains the answer to your question, especially if you take into account that $Lleft(1,left(fracDcdotright)right)$ can be estimated unconditionally (i.e. without GRH):
$$|D|^-varepsilonll_varepsilon Lleft(1,left(tfracDcdotright)right)ll log|D|.$$
The lower bound is ineffective (i.e. we don't know the implied constant), and it is due to Siegel (1934). The upper bound is effective and older. Both bounds are explained in Montgomery-Vaughan: Multiplicative number theory I.
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
$endgroup$
add a comment |
$begingroup$
My answer to this MO question contains the answer to your question, especially if you take into account that $Lleft(1,left(fracDcdotright)right)$ can be estimated unconditionally (i.e. without GRH):
$$|D|^-varepsilonll_varepsilon Lleft(1,left(tfracDcdotright)right)ll log|D|.$$
The lower bound is ineffective (i.e. we don't know the implied constant), and it is due to Siegel (1934). The upper bound is effective and older. Both bounds are explained in Montgomery-Vaughan: Multiplicative number theory I.
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
$endgroup$
add a comment |
$begingroup$
My answer to this MO question contains the answer to your question, especially if you take into account that $Lleft(1,left(fracDcdotright)right)$ can be estimated unconditionally (i.e. without GRH):
$$|D|^-varepsilonll_varepsilon Lleft(1,left(tfracDcdotright)right)ll log|D|.$$
The lower bound is ineffective (i.e. we don't know the implied constant), and it is due to Siegel (1934). The upper bound is effective and older. Both bounds are explained in Montgomery-Vaughan: Multiplicative number theory I.
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
$endgroup$
My answer to this MO question contains the answer to your question, especially if you take into account that $Lleft(1,left(fracDcdotright)right)$ can be estimated unconditionally (i.e. without GRH):
$$|D|^-varepsilonll_varepsilon Lleft(1,left(tfracDcdotright)right)ll log|D|.$$
The lower bound is ineffective (i.e. we don't know the implied constant), and it is due to Siegel (1934). The upper bound is effective and older. Both bounds are explained in Montgomery-Vaughan: Multiplicative number theory I.
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
edited 24 mins ago
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
answered 3 hours ago
GH from MOGH from MO
61.9k5 gold badges158 silver badges236 bronze badges
61.9k5 gold badges158 silver badges236 bronze badges
add a comment |
add a comment |
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